Class 12 Physics NCERT REPRINT 2026-27

Chapter 6: Electromagnetic Induction

Complete concise study notes, Faraday and Henry's experiments, magnetic flux, Faraday's & Lenz's laws, motional EMF, self & mutual inductance, AC generator, solved examples 6.1–6.10, and exercises 6.1–6.8.

01 / Exam-Focused Notes

Chapter 6: Electromagnetic Induction

Complete official NCERT textbook coverage with high-precision vector graphs: Faraday & Henry Experiments, Magnetic Flux ($\Phi_B = \mathbf{B}\cdot\mathbf{A}$), Faraday's Law ($\mathcal{E} = -N d\Phi_B/dt$), Lenz's Law & Conservation of Energy, Motional EMF ($\mathcal{E} = Blv$ and $\frac{1}{2}B\omega R^2$), Mutual ($M$) & Self-Inductance ($L = \mu_0 n^2 A l$), Magnetic Energy ($U_B = \frac{1}{2}LI^2$), and AC Generator ($\mathcal{E} = NBA\omega\sin\omega t$).

100% SYLLABUS
Chapter Navigation • NCERT Sections 6.1 – 6.8
6.1-6.2Faraday & Henry Exp. 6.3-6.4Flux & Faraday's Law 6.5Lenz's Law & Energy 6.6Motional EMF 6.7Inductance (M & L) 6.8AC Generator
6.1 & 6.2
NCERT Sections

Introduction & Experiments of Faraday and Henry

Experimental Foundations

6.2.1 Three Foundational Experiments of Faraday and Henry (1830)

  • Discovery & Significance: While Oersted and Ampere showed that currents produce magnetic fields, Faraday (England) and Henry (USA) discovered the converse: changing magnetic fields induce electric currents in closed conducting loops. This phenomenon is Electromagnetic Induction (EMI).
  • Experiment 6.1 (Magnet and Coil):
    • Pushing North-pole of a bar magnet towards coil $C_1$ causes a momentary galvanometer deflection.
    • Pulling the magnet away produces opposite deflection.
    • Faster motion produces larger deflection (larger induced current).
    • Holding the magnet stationary produces zero deflection ($\Delta\Phi_B = 0$).
    • Core Conclusion: It is the relative motion between the magnet and coil that induces current.
  • Experiment 6.2 (Current-carrying Coil and Test Coil):
    • Bar magnet is replaced by a primary coil $C_2$ carrying steady current.
    • Moving $C_2$ towards or away from test coil $C_1$ induces current in $C_1$. Relative motion between coils induces emf.
  • Experiment 6.3 (Stationary Coils with Tapping Key):
    • Two stationary coils $C_1$ and $C_2$. Pressing tapping key $K$ in $C_2$ produces a momentary deflection in $C_1$. Releasing $K$ produces an opposite momentary deflection. Continuous steady current produces zero deflection.
    • Inserting a soft iron rod dramatically increases the induced deflection.
    • Core Conclusion: Relative motion is not mandatory; time-varying magnetic field (changing flux) is the fundamental requirement for induction.
Example 6.1

Enhancing & Demonstrating Induced Currents

In Experiment 6.2: (a) What would you do to obtain a large deflection in the galvanometer? (b) How would you demonstrate the presence of induced current in the absence of a galvanometer?

(a) Large Deflection: (i) Insert a high-permeability soft iron core inside coil $C_2$, (ii) Connect $C_2$ to a high-voltage battery to increase primary current, (iii) Move $C_2$ rapidly towards $C_1$ to maximize $d\Phi_B/dt$.
(b) Demonstration without Galvanometer: Replace the galvanometer with a small low-voltage torch bulb; rapid relative motion between coils induces sufficient current to light the filament.
6.3 & 6.4
NCERT Sections

Magnetic Flux & Faraday's Law of Induction

Mathematical Formulation

6.3.1 Magnetic Flux ($\Phi_B$) & Faraday's Law of EMI

  • Magnetic Flux Definition: For a planar area $\mathbf{A}$ in uniform magnetic field $\mathbf{B}$: $$\Phi_B = \mathbf{B}\cdot\mathbf{A} = B A \cos\theta \quad (\text{SI Unit: Weber (Wb)} \equiv \text{T m}^2, \text{ Scalar})$$ For non-uniform fields: $\Phi_B = \int \mathbf{B}\cdot d\mathbf{A}$.
  • Faraday's Law of Electromagnetic Induction: The magnitude of the induced electromotive force ($\mathcal{E}$) in a circuit is equal to the time rate of change of magnetic flux through the circuit:
$$\mathcal{E} = -\frac{d\Phi_B}{dt} \qquad \text{and for } N \text{ turns: } \quad \mathcal{E} = -N \frac{d\Phi_B}{dt}$$
  • Three Independent Methods to Vary Magnetic Flux:
    1. Varying magnetic field magnitude $B(t)$ (e.g. AC solenoids, moving magnets).
    2. Changing effective surface area $A(t)$ (e.g. shrinking, stretching, sliding rod).
    3. Changing relative orientation angle $\theta(t) = \omega t$ (e.g. rotating coil in AC Generator).
Example 6.2

Induced EMF in Square Loop in Decreasing Field

A square loop of side $10\text{ cm}$ and resistance $0.5\,\Omega$ is placed vertically in the east-west plane. A field of $0.10\text{ T}$ in north-east direction ($\theta = 45^\circ$) decreases to zero in $0.70\text{ s}$ at a steady rate. Find induced EMF and current.

Initial Flux ($\Phi_i$):
$\Phi_i = B A \cos 45^\circ = 0.10 \times (0.10\text{ m})^2 \times \frac{1}{\sqrt{2}} = \frac{10^{-3}}{\sqrt{2}}\text{ Wb} \approx 0.707 \times 10^{-3}\text{ Wb}$.
Induced EMF ($\mathcal{E}$):
$\mathcal{E} = \frac{\Delta\Phi}{\Delta t} = \frac{0.707 \times 10^{-3}\text{ Wb}}{0.70\text{ s}} = 1.01 \times 10^{-3}\text{ V} \approx 1.0\text{ mV}$.
Induced Current ($I$):
$I = \frac{\mathcal{E}}{R} = \frac{1.0 \times 10^{-3}\text{ V}}{0.5\,\Omega} = 2.0 \times 10^{-3}\text{ A} = 2\text{ mA}$.
Answers: $\mathcal{E} = 1.0\text{ mV}$, $I = 2.0\text{ mA}$.
Example 6.3

Induced EMF in Rotating Coil in Earth's Magnetic Field

A circular coil of radius $10\text{ cm}$, $N = 500\text{ turns}$, $R = 2\,\Omega$ perpendicular to horizontal Earth field $B_H = 3.0 \times 10^{-5}\text{ T}$ is rotated by $180^\circ$ in $0.25\text{ s}$. Estimate induced EMF and current.

Flux Change ($\Delta\Phi$):
$\Phi_i = B_H A \cos 0^\circ = 3.0 \times 10^{-5} \times (\pi \times 0.1^2) = 3\pi \times 10^{-7}\text{ Wb}$.
$\Phi_f = B_H A \cos 180^\circ = -3\pi \times 10^{-7}\text{ Wb} \implies \Delta\Phi = 6\pi \times 10^{-7}\text{ Wb}$.
Induced EMF ($\mathcal{E}$):
$\mathcal{E} = N \frac{\Delta\Phi}{\Delta t} = 500 \times \frac{6\pi \times 10^{-7}}{0.25} = 3.77 \times 10^{-3}\text{ V} \approx 3.8\text{ mV}$.
Induced Current ($I$):
$I = \frac{\mathcal{E}}{R} = \frac{3.8 \times 10^{-3}\text{ V}}{2\,\Omega} = 1.9 \times 10^{-3}\text{ A} = 1.9\text{ mA}$.
Answers: $\mathcal{E} = 3.8\text{ mV}$, $I = 1.9\text{ mA}$.
6.5
NCERT Section

Lenz's Law and Conservation of Energy

Direction & Energy Conservation

6.5.1 Statement of Lenz's Law & Proof of Energy Conservation

Lenz's Law (H.F. Lenz, 1834): The polarity of induced EMF is such that it tends to produce an electric current which opposes the change in magnetic flux that produced it.

  • Physical Proof via Conservation of Energy:
    • When North-pole approaches a coil, induced current flows counter-clockwise (forming North pole facing magnet), repelling the magnet. Mechanical work done against repulsion is transformed into electrical energy (dissipated as Joule heat $I^2 R$).
    • Proof by Contradiction: If South pole were induced, the magnet would self-accelerate without external work, creating infinite kinetic and electrical energy from nothing — creating a perpetual motion machine that violates the First Law of Thermodynamics (Energy Conservation).
Fig 6.6: Lenz's Law: Approaching North-Pole Induces Repulsive North-Pole in Coil (Anti-Clockwise Current)
S N v → N Induced Current (Anti-Clockwise) Mechanical Work Done Against Repulsion = Electrical Energy Generated
Example 6.4

Predicting Induced Current Direction in Moving Loops

Determine induced current direction using Lenz's law for: (i) rectangular loop moving into inward $\mathbf{B}$ field, (ii) triangular loop moving out of inward $\mathbf{B}$ field, (iii) irregular loop moving out of inward $\mathbf{B}$ field.

(i) Rectangular loop entering: Inward flux increases $\implies$ induced field must be outward (counter-clockwise current $bcdab$).
(ii) Triangular loop exiting: Inward flux decreases $\implies$ induced field must be inward (clockwise current $bacb$).
(iii) Irregular loop exiting: Inward flux decreases $\implies$ induced field must be inward (clockwise current $cdabc$).
Rule: Oppose increase by generating opposite field; oppose decrease by generating reinforcing field.
Example 6.5

Conceptual Questions on Magnetic Flux vs Electric Flux

(a) Can strong static magnets generate current in stationary loop? (b) Does loop moving in electric field develop current? (c) Why is induced EMF constant for rectangular loop but variable for circular loop moving out of field? (d) Polarity of capacitor plates facing approaching magnets.

(a) Strong Static Magnets: No. Induction requires $d\Phi_B/dt \ne 0$. A constant field produces zero EMF.
(b) Motion in Electric Field: No. Varying electric flux does not produce electromagnetic induction of current in closed conductors.
(c) Rectangular vs Circular Loop: For rectangular loop, $dA/dt = l v = \text{constant} \implies \mathcal{E} = B l v$ is constant. For circular loop, $dA/dt$ varies with chord length, so $\mathcal{E}$ changes continuously.
(d) Capacitor Polarity: Approaching North poles induce counter-clockwise currents facing magnets, charging plate 'A' positive relative to plate 'B'.
6.6
NCERT Section

Motional Electromotive Force

Translational & Rotational EMF

6.6.1 Derivation of Motional EMF ($\mathcal{E} = B l v$) & Rotating Rod ($\mathcal{E} = \frac{1}{2}B\omega R^2$)

  • Translational Motional EMF (Sliding Rod):
    A conducting rod of length $l$ moves with velocity $v$ in uniform perpendicular field $B$. Area $A = l x$. $$\Phi_B = B l x \implies \mathcal{E} = -\frac{d\Phi_B}{dt} = -B l \frac{dx}{dt} = B l v \quad \text{[Eq. 6.5]}$$
  • Lorentz Force Explanation: Each free electron experiences magnetic Lorentz force $\mathbf{F}_m = -e(\mathbf{v} \times \mathbf{B})$ along the rod. Work done per unit charge: $$\mathcal{E} = \frac{W}{q} = \frac{q v B l}{q} = B l v$$
  • Induced Non-Conservative Electric Field: In stationary frame where conductor is stationary and field changes: $$\oint \mathbf{E}\cdot d\mathbf{l} = -\frac{d\Phi_B}{dt}$$
  • Rotational Motional EMF (Rotating Rod / Wheel Spokes):
    A rod of length $R$ rotates with angular speed $\omega = 2\pi\nu$ about one end hinged at center in perpendicular field $B$: $$\mathcal{E} = \int_0^R B v(r)\,dr = \int_0^R B (\omega r)\,dr = \frac{1}{2} B \omega R^2$$
$$\text{Translational: } \mathcal{E} = B l v, \qquad \text{Rotational: } \mathcal{E} = \frac{1}{2} B \omega R^2 = B \pi R^2 \nu$$
Fig 6.10: Motional EMF on Sliding Conducting Rod PQ (E = B l v)
R ××××× ××××× ××××× P (+) Q (-) v → E = B l v • I = B l v / R
Example 6.6

Rotating Rod EMF in Perpendicular Field

A metallic rod of length $1\text{ m}$ rotates with frequency $50\text{ rev/s}$ about one hinged end in uniform field $B = 1\text{ T}$ parallel to axis. Find EMF developed between center and ring.

Formula: $\mathcal{E} = \frac{1}{2} B \omega R^2 = \frac{1}{2} B (2\pi \nu) R^2 = B \pi \nu R^2$.
Calculation: $\mathcal{E} = 1.0\text{ T} \times 3.1416 \times 50\text{ s}^{-1} \times (1.0\text{ m})^2 = 157\text{ V}$.
Answer: $\mathcal{E} = 157\text{ V}$.
Example 6.7

Wheel with 10 Spokes Rotating in Earth's Magnetic Field

A wheel with 10 spokes of length $0.5\text{ m}$ rotates at $120\text{ rev/min}$ in plane normal to Earth's field $H_E = 0.4\text{ G} = 0.4 \times 10^{-4}\text{ T}$. Find induced EMF between axle and rim.

Angular Speed: $\omega = 2\pi \times \frac{120}{60} = 4\pi\text{ rad s}^{-1}$.
EMF: $\mathcal{E} = \frac{1}{2} \omega B R^2 = \frac{1}{2} \times 4\pi \times (0.4 \times 10^{-4}\text{ T}) \times (0.5\text{ m})^2 = 6.28 \times 10^{-5}\text{ V}$.
Note: Number of spokes (10) is immaterial because all spokes are connected in parallel between the same two points (axle and rim).
Answer: $\mathcal{E} = 6.28 \times 10^{-5}\text{ V}$.
6.7
NCERT Section

Inductance: Mutual Inductance & Self-Inductance

Electrical Inertia & Derivations

6.7.1 Mutual Inductance ($M$), Self-Inductance ($L$) & Energy Density ($u_B$)

  • Inductance Definition: Flux linkage is proportional to current: $N \Phi_B = L I$ or $N_1 \Phi_1 = M I_2$.
    SI Unit: Henry (H) $\equiv \text{Wb A}^{-1} \equiv \text{V s A}^{-1}$. Dimensional formula: $[M L^2 T^{-2} A^{-2}]$.
  • Mutual Inductance of Two Co-axial Solenoids:
    For inner solenoid $S_1$ ($n_1, r_1$) and outer solenoid $S_2$ ($n_2, r_2$): $$M_{12} = M_{21} = \mu_0 n_1 n_2 \pi r_1^2 l \qquad \text{or with core } \mu_r: \quad M = \mu_0 \mu_r n_1 n_2 \pi r_1^2 l$$
  • Self-Inductance of a Long Solenoid:
    Magnetic field $B = \mu_0 n I$, total flux $N\Phi_B = (n l) (\mu_0 n I) A = \mu_0 n^2 A l I$:
$$L = \mu_0 n^2 A l = \mu_0 \frac{N^2}{l} A, \qquad \mathcal{E}_{\text{back}} = -L \frac{dI}{dt}$$
  • Magnetic Energy Stored in Inductor:
    Work done against back EMF in building current $I$: $$W = \int_0^I L I' dI' = \frac{1}{2} L I^2$$ Analogy: $L$ represents electrical inertia (analogous to mass $m$ in kinetic energy $\frac{1}{2}mv^2$).
  • Magnetic Energy Density ($u_B$): Energy per unit volume inside solenoid:
$$u_B = \frac{U_B}{\text{Volume}} = \frac{B^2}{2\mu_0} \quad (\text{SI Unit: }\text{J m}^{-3})$$
Example 6.8

Mutual Inductance of Concentric Circular Coils

Two concentric coplanar circular coils have radii $r_1$ and $r_2$ ($r_1 \ll r_2$). Find mutual inductance $M_{12}$.

Field at center due to outer coil $S_2$: $B_2 = \frac{\mu_0 I_2}{2 r_2}$.
Flux linked with inner coil $S_1$: $\Phi_1 = B_2 (\pi r_1^2) = \frac{\mu_0 \pi r_1^2}{2 r_2} I_2$.
Mutual Inductance: $M_{12} = \frac{\Phi_1}{I_2} = \frac{\mu_0 \pi r_1^2}{2 r_2}$.
Answer: $M_{12} = M_{21} = \frac{\mu_0 \pi r_1^2}{2 r_2}$.
Example 6.9

Magnetic Energy Stored & Energy Density in Solenoid

(a) Obtain expression for magnetic energy stored in solenoid in terms of $B, A, l$. (b) Compare with electrostatic energy in capacitor.

(a) Energy: $U_B = \frac{1}{2} L I^2 = \frac{1}{2}(\mu_0 n^2 A l)\left(\frac{B}{\mu_0 n}\right)^2 = \frac{B^2}{2\mu_0} (A l)$.
(b) Energy Density Comparison: Magnetic energy density $u_B = \frac{B^2}{2\mu_0}$ is strictly proportional to $B^2$, analogous to electrostatic energy density $u_E = \frac{1}{2}\varepsilon_0 E^2$ in capacitor.
6.8
NCERT Section

AC Generator

Power Generation & Derivation

6.8.1 Working Principle, Mathematical Derivation & Structure

  • Principle: Converts mechanical energy into alternating electrical energy by rotating a coil of area $A$ in uniform magnetic field $B$ ($\theta = \omega t$).
  • Derivation of Instantaneous EMF:
    Magnetic flux at time $t$: $\Phi_B(t) = B A \cos(\omega t)$.
    By Faraday's law for $N$ turns: $$\mathcal{E} = -N \frac{d\Phi_B}{dt} = -N B A \frac{d}{dt}[\cos(\omega t)] = N B A \omega \sin(\omega t)$$
$$\mathcal{E}(t) = \mathcal{E}_0 \sin(\omega t) = \mathcal{E}_0 \sin(2\pi\nu t), \qquad \mathcal{E}_0 = N B A \omega = N B A (2\pi\nu)$$
  • Commercial Power Generation: Frequency $\nu = 50\text{ Hz}$ in India, $60\text{ Hz}$ in USA. In large power plants (hydro/thermal/nuclear), the armature is kept stationary and electromagnets are rotated.
Example 6.10

Bicycle Generator Maximum Voltage

A bicycle pedal is attached to a 100-turn coil of area $0.10\text{ m}^2$ rotating at $0.5\text{ rev/s}$ in $B = 0.01\text{ T}$. Find maximum voltage generated.

Formula: $\mathcal{E}_0 = N B A (2\pi\nu)$.
Calculation: $\mathcal{E}_0 = 100 \times 0.01\text{ T} \times 0.10\text{ m}^2 \times (2 \times 3.1416 \times 0.5\text{ s}^{-1}) = 0.314\text{ V}$.
Answer: $\mathcal{E}_0 = 0.314\text{ V}$.
02 / Interactive Assessment

Concept Check & Board Mastery Quiz

15 targeted MCQs covering Faraday's laws, Lenz's law, motional EMF, self & mutual inductance, and AC generator with in-depth explanations.

15 MCQS
0 / 15 Answered
Score: 0 (0%)
03 / Textbook Solutions

NCERT Exercises 6.1 – 6.8

Complete stepwise solutions for every textbook exercise question in NCERT Class 12 Physics Chapter 6 Electromagnetic Induction Reprint 2026-27.

8 QUESTIONS
Ex 6.1 Direction of Induced Current (Lenz's Law)

6.1 Predict the direction of induced current in the situations described by Figs. 6.15(a) to (f):
(a) South pole of bar magnet approaching coil side 'q'.
(b) North pole receding from coil 1; South pole approaching coil 2.
(c) Tapping key closed (current growing in neighbouring circuit).
(d) Rheostat setting being changed (rheostat resistance decreasing).
(e) Tapping key released (current decaying).
(f) Straight wire current decreasing.

(a) Approaching South Pole: The approaching South pole induces a South pole at face 'q' to repel it $\implies$ clockwise current from viewer's side, flowing along $q \to r \to p \to q$.
(b) Receding North / Approaching South: (i) Left coil attracts receding North by inducing South pole $\implies$ current along $p \to r \to q \to p$. (ii) Right coil repels approaching South by inducing South pole $\implies$ current along $y \to z \to x \to y$.
(c) Tapping Key Pressed: Current in right coil grows from zero $\implies$ flux through left coil increases $\implies$ induced current opposes growth, flowing along $y \to z \to x \to y$.
(d) Rheostat Resistance Decreased: Current in primary circuit increases $\implies$ flux increases $\implies$ induced current opposes growth, flowing along $z \to y \to x \to z$.
(e) Tapping Key Released: Current in primary circuit decays to zero $\implies$ flux decreases $\implies$ induced current reinforces decay, flowing along $x \to r \to y \to x$.
(f) Current Decreasing in Straight Wire: Magnetic field lines are in the plane of the loop $\implies$ magnetic flux linked with loop is strictly zero at all times $\implies$ no induced current in the loop.
Ex 6.2 Lenz's Law on Deforming Loops

6.2 Use Lenz's law to determine the direction of induced current in the situations described by Fig. 6.16:
(a) A wire of irregular shape turning into a circular shape in an outward magnetic field ($\odot$);
(b) A circular loop being deformed into a narrow straight wire in an inward magnetic field ($\otimes$).

(a) Irregular to Circular (Outward Field $\odot$):
Circle encloses maximum area for a given perimeter $\implies$ area of loop increases $\implies$ outward magnetic flux increases.
To oppose this increase, induced current must produce an inward magnetic field ($\otimes$) $\implies$ induced current flows in clockwise direction along $a \to d \to c \to b \to a$.
(b) Circular to Straight Wire (Inward Field $\otimes$):
Deforming into a narrow wire decreases area to zero $\implies$ inward magnetic flux decreases.
To oppose this decrease, induced current must produce an inward magnetic field ($\otimes$) $\implies$ induced current flows in clockwise direction along $a \to b \to c \to d \to a$.
Ex 6.3 Induced EMF in Solenoid Interior Loop

6.3 A long solenoid with 15 turns per cm has a small loop of area $2.0\text{ cm}^2$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from $2.0\text{ A}$ to $4.0\text{ A}$ in $0.1\text{ s}$, what is the induced emf in the loop while the current is changing?

Induced EMF $\mathcal{E} = 7.5 \times 10^{-6}\text{ V} = 7.5\,\mu\text{V}$.
Given:
$n = 15\text{ turns/cm} = 1500\text{ turns/m}$, $A = 2.0\text{ cm}^2 = 2.0 \times 10^{-4}\text{ m}^2$,
$\frac{dI}{dt} = \frac{4.0 - 2.0}{0.1} = \frac{2.0}{0.1} = 20\text{ A s}^{-1}$, $\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}$.
Formula:
Field inside solenoid $B = \mu_0 n I \implies \Phi = B A = \mu_0 n I A$.
$$\mathcal{E} = \frac{d\Phi}{dt} = \mu_0 n A \frac{dI}{dt}$$
Calculation:
$$\mathcal{E} = (4\pi \times 10^{-7}) \times 1500 \times (2.0 \times 10^{-4}) \times 20$$ $$\mathcal{E} = (1.2566 \times 10^{-6}) \times 1500 \times 4.0 \times 10^{-3} = 7.54 \times 10^{-6}\text{ V} \approx 7.5 \times 10^{-6}\text{ V}$$
Ex 6.4 Rectangular Loop Moving Out of Field

6.4 A rectangular wire loop of sides $8\text{ cm}$ and $2\text{ cm}$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3\text{ T}$ directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is $1\text{ cm s}^{-1}$ in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

(a) $\mathcal{E} = 0.24\text{ mV}$, lasts for $2.0\text{ s}$ • (b) $\mathcal{E} = 0.06\text{ mV}$, lasts for $8.0\text{ s}$.
Given: $l = 8\text{ cm} = 0.08\text{ m}$, $b = 2\text{ cm} = 0.02\text{ m}$, $B = 0.3\text{ T}$, $v = 1\text{ cm s}^{-1} = 0.01\text{ m s}^{-1}$.
(a) Velocity normal to longer side ($l = 8\text{ cm}$ cuts field lines):
$\mathcal{E} = B l v = 0.3\text{ T} \times 0.08\text{ m} \times 0.01\text{ m s}^{-1} = 2.4 \times 10^{-4}\text{ V} = 0.24\text{ mV}$.
Distance moved to completely exit $= b = 2\text{ cm} \implies t = \frac{b}{v} = \frac{2\text{ cm}}{1\text{ cm s}^{-1}} = 2.0\text{ s}$.
(b) Velocity normal to shorter side ($b = 2\text{ cm}$ cuts field lines):
$\mathcal{E} = B b v = 0.3\text{ T} \times 0.02\text{ m} \times 0.01\text{ m s}^{-1} = 0.6 \times 10^{-4}\text{ V} = 0.06\text{ mV}$.
Distance moved to completely exit $= l = 8\text{ cm} \implies t = \frac{l}{v} = \frac{8\text{ cm}}{1\text{ cm s}^{-1}} = 8.0\text{ s}$.
Ex 6.5 Rotating Metallic Rod EMF

6.5 A $1.0\text{ m}$ long metallic rod is rotated with an angular frequency of $400\text{ rad s}^{-1}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of $0.5\text{ T}$ parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Induced EMF $\mathcal{E} = 100\text{ V}$.
Formula: Motional EMF of a rotating rod: $$\mathcal{E} = \frac{1}{2} B \omega R^2$$
Given: $B = 0.5\text{ T}$, $\omega = 400\text{ rad s}^{-1}$, $R = 1.0\text{ m}$.
Calculation: $$\mathcal{E} = \frac{1}{2} \times 0.5\text{ T} \times 400\text{ rad s}^{-1} \times (1.0\text{ m})^2 = 0.25 \times 400 = 100\text{ V}$$
Ex 6.6 Horizontal Wire Falling in Earth's Magnetic Field

6.6 A horizontal straight wire $10\text{ m}$ long extending from east to west is falling with a speed of $5.0\text{ m s}^{-1}$, at right angles to the horizontal component of the earth’s magnetic field, $0.30 \times 10^{-4}\text{ Wb m}^{-2}$.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?

(a) $\mathcal{E} = 1.5 \times 10^{-3}\text{ V} = 1.5\text{ mV}$ • (b) Directed from West to East • (c) Western end is at higher potential.
(a) Magnitude of Induced EMF:
$$\mathcal{E} = B_H l v = (0.30 \times 10^{-4}\text{ T}) \times 10\text{ m} \times 5.0\text{ m s}^{-1} = 1.5 \times 10^{-3}\text{ V} = 1.5\text{ mV}$$
(b) Direction of Induced EMF:
By Fleming's Right-Hand Rule: Field is Northward ($\mathbf{B}$), Motion is Downward ($\mathbf{v}$), Induced Current / EMF direction points from West to East.
(c) Higher Potential End:
Free electrons experience magnetic Lorentz force $\mathbf{F}_m = -e(\mathbf{v} \times \mathbf{B})$ directed towards the East, accumulating negative charge at the East end. Hence, the Western end is at higher electrical potential.
Ex 6.7 Self-Inductance from Decay of Current

6.7 Current in a circuit falls from $5.0\text{ A}$ to $0.0\text{ A}$ in $0.1\text{ s}$. If an average emf of $200\text{ V}$ is induced, give an estimate of the self-inductance of the circuit.

Self-Inductance $L = 4.0\text{ H}$.
Formula: Back EMF induced in an inductor: $$\mathcal{E} = -L \frac{dI}{dt} \implies L = \frac{\mathcal{E}}{|\Delta I / \Delta t|}$$
Given: $\mathcal{E} = 200\text{ V}$, $\Delta I = 5.0 - 0.0 = 5.0\text{ A}$, $\Delta t = 0.1\text{ s} \implies |\Delta I / \Delta t| = 50\text{ A s}^{-1}$.
Calculation: $$L = \frac{200\text{ V}}{50\text{ A s}^{-1}} = 4.0\text{ H}$$
Ex 6.8 Mutual Flux Linkage Change

6.8 A pair of adjacent coils has a mutual inductance of $1.5\text{ H}$. If the current in one coil changes from $0$ to $20\text{ A}$ in $0.5\text{ s}$, what is the change of flux linkage with the other coil?

Change of Flux Linkage $\Delta(N_2 \Phi_2) = 30\text{ Wb}$.
Formula: Total flux linkage in mutual induction: $$N_2 \Phi_2 = M I_1 \implies \Delta(N_2 \Phi_2) = M \Delta I_1$$
Given: $M = 1.5\text{ H}$, $\Delta I_1 = 20 - 0 = 20\text{ A}$.
Calculation: $$\Delta(N_2 \Phi_2) = 1.5\text{ H} \times 20\text{ A} = 30\text{ Wb}$$ (Note: The time duration $0.5\text{ s}$ is not needed since flux linkage change depends only on current change).
04 / Rapid Reference

Chapter Summary & Points to Ponder

9 core summary principles, Physical Quantities dimension and unit table, and official NCERT Points to Ponder for Chapter 6.

100% SYLLABUS

Magnetic Flux Definition

ΦB = B · A = BA cos θ (Weber, Wb)

Scalar quantity; depends on field strength $B$, area $A$, and angle $\theta$ with area normal.

Faraday's Law of EMI

Ε = −N dΦB / dt

Induced EMF is directly proportional to the time rate of change of magnetic flux.

Lenz's Law & Energy

Induced current opposes flux change

Direct consequence of the conservation of energy; mechanical work transforms to electrical heat.

Translational Motional EMF

Ε = B l v

Conductor of length $l$ sliding perpendicular to uniform field $B$ with speed $v$.

Rotational Motional EMF

Ε = ½ B ω R² = B π R² ν

Rod/spoke rotating about one hinged end in perpendicular magnetic field.

Mutual Inductance of Solenoids

M = μ₀ n₁ n₂ π r₁² l

Reciprocity relation holds generally: $M_{12} = M_{21} = M$.

Self-Inductance of Solenoid

L = μ₀ n² A l • Ε_back = −L dI/dt

Measures electrical inertia; magnetic potential energy $U_B = \frac{1}{2} L I^2$.

Magnetic Energy Density

u_B = B² / (2μ₀) (J m⁻³)

General energy density stored per unit volume in any magnetic field.

AC Generator Equation

Ε(t) = NBAω sin(ωt)

Peak voltage $\mathcal{E}_0 = N B A (2\pi\nu)$; converts mechanical energy to AC electricity.

NCERT Official Table

Physical Quantities, Symbols, Dimensions & Units

Physical Quantity Symbol Dimensions SI Unit Governing Equation
Magnetic Flux ΦB $[M L^2 T^{-2} A^{-1}]$ Wb (weber $\equiv$ T m$^2$) $\Phi_B = \mathbf{B}\cdot\mathbf{A} = B A \cos\theta$
Induced EMF Ε $[M L^2 T^{-3} A^{-1}]$ V (volt) $\mathcal{E} = -N \frac{d\Phi_B}{dt}$
Mutual Inductance M $[M L^2 T^{-2} A^{-2}]$ H (henry $\equiv$ V s A$^{-1}$) $\mathcal{E}_1 = -M_{12}\frac{dI_2}{dt}$
Self Inductance L $[M L^2 T^{-2} A^{-2}]$ H (henry $\equiv$ V s A$^{-1}$) $\mathcal{E} = -L \frac{dI}{dt}$
Magnetic Energy Density u_B $[M L^{-1} T^{-2}]$ J m$^{-3}$ $u_B = \frac{B^2}{2\mu_0}$
NCERT Official

Points to Ponder

  1. Intimate Relation: Electricity and magnetism are intimately unified. While Oersted and Ampere established that moving charges create magnetic fields, Faraday and Henry proved that changing magnetic fields generate electric currents.
  2. Open vs Closed Circuits: In a closed circuit, induced currents physically flow and produce observable magnetic opposition (Lenz's law). In an open circuit, an induced EMF $\mathcal{E} = -d\Phi_B/dt$ still develops across the terminals, maintaining electric charge separation without steady current.
  3. Lorentz Force & Relativity: Motional EMF $\mathcal{E} = B l v$ can be derived directly from Lorentz force $q(\mathbf{v} \times \mathbf{B})$. When the conductor is stationary and field varies, the force on charges arises from an induced non-conservative electric field $\mathbf{E}$. Moving charges in a static field and static charges in a time-varying field present a symmetric manifestation that hinted at Einstein's Special Theory of Relativity.
05 / Practice Tests

3-Tier Practice Tests

Level 1 (Foundation), Level 2 (Application), and Level 3 (Challenge). Select options and submit to evaluate your preparation.

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